Hey,
Answer:
<em>AgCl</em><em> </em>(Silver Chloride)
Explanation:
sodium chloride + silver nitrate → sodium nitrate+ silver chloride
Hope it helps :)
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Answer:
9 protons, 10 neutrons, and 9 electrons.
Explanation:
The particles of the nucleus of an atom of Fluorine-19 is
9 protons, 10 neutrons, and 9 electrons.
The new temperature of the piece of lead will be 37.5°C
SPECIFIC HEAT CAPACITY:
- The amount of heat absorbed or released by a substance can be calculated by using the following formula:
Q = m × c × ∆T
Where;
Q = quantity of heat absorbed or released (J)
m = mass of substance (g)
c = specific heat capacity
∆T = change in temperature (°C)
According to this question,
Q = 1.90J
m = 1.29g
T1 = 26°C
T2 = ?
c = 0.128 J/g°C
1.90 = 1.29 × 0.128 × (T2- 26°C)
1.90 = 0.165 (T2 - 26°C)
1.90 = 0.165T2 - 4.29
1.90 + 4.29 = 0.165T2
6.19 = 0.165T2
T2 = 37.5°C
Therefore, the new temperature of the piece of lead will be 37.5°C.
Learn more: brainly.com/question/20514651?referrer=searchResults
Explanation:
The reaction equation will be as follows.
Using bond energies, expression for calculating the value of 
is as follows.
On reactant side, from 
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 6
From 
; Cl-Cl bonds = 1
On product side, from 
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 5
C-Cl bonds = 1
From HCl; H-Cl bonds = 1
Hence, using the bond energies we will calculate the enthalpy of reaction as follows.
=![[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]](https://tex.z-dn.net/?f=%5B%281%20%5Ctimes%20348%20kJ%2Fmol%29%20%2B%20%286%20%5Ctimes%20414%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20242%20kJ%2Fmol%29%5D%20-%20%5B%281%20%5Ctimes%20348%20kJ%2Fmol%29%20%2B%20%285%20%5Ctimes%20414%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20327%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20431%20kJ%2Fmol%29%5D)
= -102 kJ/mol
Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.
